bn ơi hình như đề bài thiếu cái j đó

\(\frac{4x+3}{2x+1}=\frac{2x+1}{x+2}\)

\(\Rightarrow\frac{\left(4x+2\right)+1}{2x+1}=\frac{\left(2x+4\right)-3}{x+2}\)

\(\Rightarrow\frac{2.\left(2x+1\right)}{2x+1}+\frac{1}{2x+1}=\frac{2.\left(x+2\right)}{x+2}-\frac{3}{x+2}\)

\(\Rightarrow2+\frac{1}{2x+1}=2-\frac{3}{x+2}\)

\(\Rightarrow\frac{1}{2x+1}=\frac{3}{x+2}\)

\(\Rightarrow x+2=3.\left(2x+1\right)\)

\(\Rightarrow x+2=6x+3\)

\(\Rightarrow x-6x=3-2\)

\(\Rightarrow-5.x=1\Rightarrow x=-\frac{1}{5}\)

Nhầm 1 chút nha bạn.

\(\Rightarrow\frac{1}{2x+1}=-\frac{3}{x+2}\)

\(\Rightarrow x+2=-3.\left(2x+1\right)\)

\(\Rightarrow x+2=-6x-3\)

\(\Rightarrow6x+x=-3-2\)

\(\Rightarrow7x=-5\Rightarrow x=-\frac{5}{7}\)

Theo bài ra ta có :

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Leftrightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\)

Áp dụng tính chất của dãy tỉ số bằng nhau , ta có :

\(\frac{2x}{3}=\frac{3y}{4}=\frac{4z}{5}\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}\Rightarrow\frac{x+y+z}{18+16+15}=\frac{49}{49}=1\)

\(\frac{x}{18}=1\Rightarrow x=1\)

\(\frac{y}{16}=1\Rightarrow y=16\)

\(\frac{z}{15}=1\Rightarrow z=15\)

Vậy x = 18 , y = 16 , z = 15

\(2x=3y=5z\Rightarrow\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(2x=3y=5z\Rightarrow\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x-y+z}{\frac{1}{2}-\frac{1}{3}+\frac{1}{5}}=-\frac{33}{\frac{11}{30}}=-\frac{1}{10}\)

\(2x=-\frac{1}{10}\Rightarrow x=-\frac{1}{10}:2=-\frac{1}{20}\)

\(3y=-\frac{1}{10}\Rightarrow y=-\frac{1}{10}:3=-\frac{1}{30}\)

\(5z=-\frac{1}{10}\Rightarrow z=-\frac{1}{10}:5=-\frac{1}{50}\)

\(33^{44}=33^{4.11}=\left(33^4\right)^{11}=1185921^{11}\)

\(44^{33}=44^{3.11}=\left(44^3\right)^{11}=85184^{11}\)

Vì \(1185921^{11}>85184^{11}\)

\(\Rightarrow33^{44}>44^{33}\)